# Space of continuous functions with mapping to $\mathbb{R}^n$ with compact support are dense in $F(X,\mathbb{R}^{n-1})$

I have the following statements

At first, theorem 1:
Let
$$A$$ be a complete metric space,
$$B$$ a separable metric space,
$$M$$ in $$A\times B$$ open and dense set, then

the room $$A$$ contains a dense set $$N$$ with the property, that for every point $$a$$ of $$N$$ in $$B$$ exists a dense set of points $$y$$, such that all points $$(a,y)$$ belongs to $$M$$. End theorem 1.

Let $$F_n(X):=\{f:X\rightarrow \mathbb{R}^n \;|\;f\;continuous \}$$ the space of continuous functions with mapping to $$\mathbb{R}^n$$ on a compact metric space $$X$$ provided with sup-metric and
$$G_n(X):=\{f:X\rightarrow \mathbb{R}^n \;|\; f\; continuous,\; f(r)\neq 0, \; \forall \; x\in X\}$$ the space of continuous functions with mapping to $$\mathbb{R}^n$$ without functions with zeros, on a compact metric space $$X$$ provided with sup-metric.

So $$F_n(X)$$ is a complete, separable metric space, $$G_n(X)$$ is open in $$F_n(X)$$ and we can write $$F_n(X)$$ as $$F_1(X)\times F_{n-1}(X)$$.

So if $$G_n(X)$$ is dense in $$F_n(X)$$ we can use theorem 1.
Therefore there exists a everywhere dense of functions $$f\in F_1(X)$$ such that:
a) For all functions $$\varphi\in F_{n-1}$$ who combined with $$f$$ yield an element $$(f,\varphi)\in G_n(X)$$, form a dense set in $$F_{n-1}$$

If we define for $$f\in F_1(X)$$ with $$E(f)$$ as set of points $$x\in X$$ with $$f(x)=0$$, then for $$\varphi\in F_{n-1}$$ is the relation $$(f,\varphi)\in G_n(X)$$ equivalent, that no point of $$E(f)$$ will be mapped through $$\varphi$$ in the point $$(0,0,\ldots, 0)\in \mathbb{R}^{n-1}$$.

I don't understand why $$\varphi(x)$$ with $$x\in E(f)$$ can't be mapped to the zero vector. Because, let us start with $$F_1(X)\times F_1(X)$$. Have a look a my picture (sorry for ugly writing). Visualisation of my imagination

so if we take one constant $$f$$ in $$G_2(X)$$ e.g. $$f(x)=1$$, then $$\varphi$$ can be mapped to $$0$$ in $$G_2$$ (because the constraint that functions with zeros in $$G_2$$ are not allowed are not broken) and therefore it is allowed to use $$\varphi(x)$$ with $$x\in E(f)$$ if $$f$$ is not a function with zero.

Can anyone help me to find my logic mistake? Thanks

Therefore there exists a everywhere dense of functions $$f\in F_1(X)$$ such that:
a) For all functions $$\varphi\in F_{n-1}$$ who combined with $$f$$ yield an element $$(f,\varphi)\in G_n(X)$$, form a dense set in $$F_{n-1}$$

That is there exists a dense set $$N$$ of $$F_1(x)$$ such that for each $$f\in N$$ the set $$G_f=\{\varphi\in F_{n-1}(X):(f,\varphi)\in G_{n}(X)\}$$ is dense. Note that the set $$G_f$$ depends on the function $$f$$ and it may be not the same for distinct $$f\in F_1(X)$$.

If we define for $$f\in F_1(X)$$ with $$E(f)$$ as set of points $$x\in X$$ with $$f(x)=0$$, then for $$\varphi\in F_{n-1}$$ is the relation $$(f,\varphi)\in G_n(X)$$ equivalent, that no point of $$E(f)$$ will be mapped through $$\varphi$$ in the point $$(0,0,\ldots, 0)\in \mathbb{R}^{n-1}$$.

Right.

I don't understand why $$\varphi(x)$$ with $$x\in E(f)$$ can't be mapped to the zero vector.

Note, that we still speak about pairs $$(f,\varphi)\in G_n(X)$$. And if $$x\in E(f)$$ and $$\varphi(x)=0$$ then $$(f(x),\varphi(x))=0$$, which contradicts $$(f,\varphi)\in G_n(X)$$.

so if we take one constant $$f$$ in $$G_2(X)$$ e.g. $$f(x)=1$$, then $$\varphi$$ can be mapped to $$0$$ in $$G_2$$ (because the constraint that functions with zeros in $$G_2$$ are not allowed are not broken) and therefore it is allowed to use $$\varphi(x)$$ with $$x\in E(f)$$ if $$f$$ is not a function with zero.

We are allowed to use $$\varphi$$ means $$(f,\varphi)\in G_n(X)$$ for a given fixed $$f$$. This doesn’t imply that we are allowed to use $$\varphi$$ with a different function, say $$f’\in F_1(X)$$. For a particular choice of $$f$$, when $$f(x)=1$$ for each $$x\in X$$, a set $$E(f)=\{x\in X :f(x)=0\}$$ is empty. Then $$G_f=\{\varphi\in F_{n-1}(X):(f,\varphi)\in G_{n}(X)\}=F_{n-1}(X)$$. But this does not mean that $$G_{f’}=F_{n-1}(X)$$ for any function $$f’\in F_1(X).$$

Remark that in this case also $$E(f)=\{x\in X :f(x)=0\}$$ is empty, so we cannot "use $$\varphi(x)$$ with $$x\in E(f)$$", because there is no such $$x$$. But if $$f’\in F_1(X)$$ is an other function then we are allowed to use a function $$\varphi\in F_{n-1}$$ with the function $$f’$$, that is $$(f,\varphi)\in G_n(X)$$ iff $$\varphi(x)\ne 0$$ for each $$x\in E(f’)$$.

PS. It seems that the picture is not needed.

• And what about if only one $f$ is constant e.g. $f=1$ and the other have zeros. So one can write $(1,\varphi)$ with $\varphi$ abitrary. Then it did not contradicts $(f,\varphi)\in G_n(X)$? – GeoRie Dec 6 '18 at 14:28
• @GeoRie Looking for $(f,\varphi)\in G_n(X)$, we cannot pair arbitrary $f$ with arbitrary $\varphi$, we need a restriction $\varphi(x)\ne(0,0,\dots,0)\in\Bbb R^{n-1}$ for each $x\in E(f)$, right? – Alex Ravsky Dec 6 '18 at 16:09
• Yes. That's my question. Why we need a restriction to $\varphi$ if we combine it with those $f$ who did not have zeros. Then we don't need a restriction for $\varphi$? Thrn, if we choose such $f$, $\varphi$ can be mapped to zero in $\mathbb{R}$? – GeoRie Dec 7 '18 at 9:21
• Do you understand my thoughts? Are my thoughts wrong? Could you explain it more exact, please. Thanks a lot. – GeoRie Dec 9 '18 at 11:40
• @GeoRie I expanded my answer. – Alex Ravsky Dec 9 '18 at 12:16